UVALive2158 POJ1401 HDU1124 ZOJ2024 Factorial【分析思维】

Factorial

Time Limit: 1500MS		Memory Limit: 65536K
Total Submissions: 16314		Accepted: 10052

Description

The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this term gave the name to the cellular phone) and every phone connects to the BTS with the strongest signal (in a little simplified view). Of course, BTSes need some attention and technicians need to check their function periodically. 

ACM technicians faced a very interesting problem recently. Given a set of BTSes to visit, they needed to find the shortest path to visit all of the given points and return back to the central company building. Programmers have spent several months studying this problem but with no results. They were unable to find the solution fast enough. After a long time, one of the programmers found this problem in a conference article. Unfortunately, he found that the problem is so called "Travelling Salesman Problem" and it is very hard to solve. If we have N BTSes to be visited, we can visit them in any order, giving us N! possibilities to examine. The function expressing that number is called factorial and can be computed as a product 1.2.3.4....N. The number is very high even for a relatively small N. 

The programmers understood they had no chance to solve the problem. But because they have already received the research grant from the government, they needed to continue with their studies and produce at least some results. So they started to study behaviour of the factorial function. 

For example, they defined the function Z. For any positive integer N, Z(N) is the number of zeros at the end of the decimal form of number N!. They noticed that this function never decreases. If we have two numbers N1 < N2, then Z(N1) <= Z(N2). It is because we can never "lose" any trailing zero by multiplying by any positive number. We can only get new and new zeros. The function Z is very interesting, so we need a computer program that can determine its value efficiently. 

Input

There is a single positive integer T on the first line of input. It stands for the number of numbers to follow. Then there is T lines, each containing exactly one positive integer number N, 1 <= N <= 1000000000.

Output

For every number N, output a single line containing the single non-negative integer Z(N).

Sample Input

6
3
60
100
1024
23456
8735373

Sample Output

0
14
24
253
5861
2183837

Source

Central Europe 2000

Regionals 2000 >> Europe - Central

问题链接：UVALive2158 POJ1401 HDU1124 ZOJ2024 Factorial。

问题简述：参见上文。

问题分析：

计算n!的末尾0个数，依赖于n!中各个因子的中5的因子的个数。因为2*5=10，那么0的个数依赖于其因子中的2的个数和5的个数。2的个数自然比5的个数多很多，就只需要知道5的个数即可。

令f(x)表示正整数x末尾所含有的“0”的个数, g(x)表示正整数x的因式分解中因子“5”的个数，那么有f(n!) = g(n!) = g(5^k * k! * a) = k + g(k!) = k + f(k!)。

程序说明：（略）

参考链接：（略）

题记：（略）

AC的C++语言程序如下：

/* UVALive2158 POJ1401 HDU1124 ZOJ2024 Factorial */

#include <iostream>
#include <stdio.h>

using namespace std;

inline int getlast0s(int n)
{
    int ans = 0;

    while(n) {
        ans += n / 5;
        n /= 5;
    }

    return ans;
}

int main()
{
    int t, n;

    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);

        printf("%d
", getlast0s(n));
    }

    return 0;
}